Question

x6 -8 by x2 - 2 . find quotient and remainder if any

Answer 1

Step-by-step explanation:

No. because whenever we divide a polynomial x6 + 2x3 + x -1 by a polynomial in x of degree 5, then we get quotient always as in linear form i.e., polynomial in x of degree 1. Let divisor = a polynomial in x of degree 5

= ax5 + bx4 + cx3 + dx2 + ex + f

quotient = x2 -1

and dividend = x6 + 2x3 + x -1

By division algorithm for polynomials,

Dividend = Divisor x Quotient + Remainder

= (ax5 + bx4 + cx3 + dx2 + ex + f)x(x2 -1) + Remainder

= (a polynomial of degree 7) + Remainder

[in division algorithm, degree of divisor > degree of remainder]

= (a polynomial of degree 7)

But dividend = a polynomial of degree 6

So, division algorithm is not satisfied.

Hence, x2 -1 is not a required quotient.

Answer 2

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No. because whenever we divide a polynomial x6 + 2x3 + x -1 by a polynomial in x of degree 5, then we get quotient always as in linear form i.e., polynomial in x of degree 1. Let divisor = a polynomial in x of degree 5

= ax5 + bx4 + cx3 + dx2 + ex + f

quotient = x2 -1

and dividend = x6 + 2x3 + x -1

By division algorithm for polynomials,

Dividend = Divisor x Quotient + Remainder

= (ax5 + bx4 + cx3 + dx2 + ex + f)x(x2 -1) + Remainder

= (a polynomial of degree 7) + Remainder

[in division algorithm, degree of divisor > degree of remainder]

= (a polynomial of degree 7)

But dividend = a polynomial of degree 6

So, division algorithm is not satisfied.

Hence, x2 -1 is not a required quotient